If $A$ and $B$ are any two events such that $P(A) = \frac{2}{5}$ and $P(A \cap B) = \frac{3}{20}$,then the conditional probability $P(A | A' \cup B')$,where $A'$ denotes the complement of $A$,is equal to:

Evaluate $P(A \cup B)$,if $2 P(A) = P(B) = \frac{5}{13}$ and $P(A|B) = \frac{2}{5}$.

$A$ couple has two children. If at least one of them is a boy,then the probability that the other is also a boy is:

If $P(A) = \frac{1}{2}$,$P(B) = \frac{1}{3}$ and $P(A \cap B) = \frac{1}{4}$,then $P(B/A) = $

$A$ bag contains $N$ balls out of which $3$ are white,$6$ are green,and the remaining $(N-9)$ balls are blue. Three balls are drawn randomly one after the other without replacement. Let $W_i, G_i$,and $B_i$ denote the events that the ball drawn in the $i^{\text{th}}$ draw is white,green,and blue,respectively. If $P(W_1 \cap G_2 \cap B_3) = \frac{2}{5N}$ and $P(B_3 \mid W_1 \cap G_2) = \frac{2}{9}$,then $N$ equals:

$A$ cubical die with faces marked $1, 2, 3, ..., 6$ is tossed such that the probability of throwing the number $t$ is proportional to $t^2$. The probability that the number $5$ has appeared,given that the number turned up is not even,is: